Answer:
All atoms heavier than barium
Explanation:
In the periodic table, elements are divided into blocks. We have the;
s- block elements
p- block elements
d- block elements
f- block elements
However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.
Hence elements heavier than barium all possess f-orbitals.
If you are talking about bonds they are influenced by electronegativity and how many valence electrons they have.
In this redox reaction, the Cu goes from oxidation state of (0) to (+2), therefore it oxidises. N in HNO₃ goes from oxidation state of (+5) to N in NO with oxidation state of (+2) and becomes reduced.
Cu acts as the reducing reagent and HNO₃ is the oxidising agent.
oxidation half reaction
Cu ---> Cu²⁺ + 2e --1)
reduction half reaction
4H⁺ + 3e + NO₃⁻ ---> NO + 2H₂O --2)
to balance the number of electrons , 1) x3 and 2) x2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 6e + 2NO₃⁻ ---> 2NO + 4H₂O
add the 2 equations
3Cu + 8H⁺ + 2NO₃⁻ ---> 3Cu²⁺ + 2NO + 4H₂O
add 6 nitrate ions to both sides to add up to 8 and form acid with 8H⁺ ions
3Cu + 8HNO₃ ---> 3Cu(NO₃)₂ + 2NO + 4H₂O
Balanced equation for the redox reaction is as follows;
3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O<span>(l)
NO has a coefficient of 2
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<span>Magnetic quantum number specify orientation of electrons in magnetic field </span>and number of electron states (orbitals) in subshells..
Magnetic quantum number (ml) specifies the orientation in space of an orbital of a given energy and shape . Magnetic quantum number divides the subshell into individual orbitals which hold the electrons, there are 2l+1
orbitals in each subshell. For example, p orbitals (and their electrons) have three orientations in spase (px, py and pz).
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
